What is the integral from 0 to 4 of (sqrt(x))/(1+x) ?

The integral from 0 to 4 of (sqrt(x))/(1+x) is 2(-arctan(2)+2)

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integral from 0 to 4 of (sqrt(x))/(1+x)

\int_{0}^{4} \frac{x}{1 + x} d x

2 (- arctan (2) + 2)

Decimal

1.78570 \dots

Solution steps

\int_{0}^{4} \frac{x}{1 + x} d x

Apply u-substitution

= \int_{0}^{2} \frac{2 u ^{2}}{1 + u ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{2} \frac{u ^{2}}{1 + u ^{2}} d u

\frac{u ^{2}}{1 + u ^{2}} = - \frac{1}{1 + u ^{2}} + 1

= 2 \cdot \int_{0}^{2} - \frac{1}{1 + u ^{2}} + 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (- \int_{0}^{2} \frac{1}{1 + u ^{2}} d u + \int_{0}^{2} 1 d u)

\int_{0}^{2} \frac{1}{1 + u ^{2}} d u = arctan (2)

\int_{0}^{2} 1 d u = 2

= 2 (- arctan (2) + 2)

What is the integral from 0 to 4 of (sqrt(x))/(1+x) ?
The integral from 0 to 4 of (sqrt(x))/(1+x) is 2(-arctan(2)+2)